If Σᵣ₌₁⁵⁰ tan ⁻¹1/2r² = p, then the value of tan p is :

Solution

Maths sir · 4 steps

Rewrite each term as a difference of two inverse tangents, and the whole sum collapses to two terms.

1. Reshape the term. Multiply top and bottom by 2: 1/2r² = 2/4r² = 2/(1 + (4r² - 1)). Now 2 = (2r+1) - (2r-1) and 4r² - 1 = (2r+1)(2r-1).

   tan⁻¹ 1/2r² = tan⁻¹((2r+1)-(2r-1))/(1+(2r+1)(2r-1)) = tan⁻¹(2r+1) - tan⁻¹(2r-1)

2. Telescope the sum. For r = 1, 2, …, 50 the terms are (tan⁻¹ 3 - tan⁻¹ 1) + (tan⁻¹ 5 - tan⁻¹ 3) + … + (tan⁻¹ 101 - tan⁻¹ 99). Every middle term cancels, leaving p = tan⁻¹ 101 - tan⁻¹ 1.
3. Take the tangent. Use tan(A - B) = (tan A - tan B)/(1 + tan A tan B) with tan A = 101 and tan B = 1. This gives tan p = (101 - 1)/(1 + 101 × 1) = 100/102.
4. So. tan p = 50/51, which is option (b).

The idea to keep. When a sum of inverse tangents appears, try to write each term as (x - y)/(1 + xy). It then splits into tan⁻¹ x - tan⁻¹ y and the sum telescopes.

Answer from the NTA final answer key. Solution by PrepWiser.